Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Let }}I = \int {\frac{{{x^{n - 1}}dx}}{{{x^{2n}} + {a^2}}}} \cr & {\text{Let }}{x^n} = t \Rightarrow n.{x^{n - 1}}dx = dt \cr & \therefore \,I = \int {\frac{1}{n}.\frac{{dt}}{{{t^2} + {a^2}}}} \cr & = \frac{1}{n}.\frac{1}{a}{\tan ^{ - 1}}\left( {\frac{t}{a}} \right) + C \cr & = \frac{1}{{na}}{\tan ^{ - 1}}\left[ {\frac{{{x^n}}}{a}} \right] + C \cr} $$

$$\eqalign{ & \int {\frac{{{{\left( {\log \,x - 1} \right)}^2}}}{{{{\left( {1 + {{\left( {\log \,x} \right)}^2}} \right)}^2}}}dx} \cr & = \int {\frac{{1 + {{\left( {\log \,x} \right)}^2} - 2\,\log \,x}}{{{{\left[ {1 + {{\left( {\log \,x} \right)}^2}} \right]}^2}}}} dx \cr & = \int {\left[ {\frac{1}{{\left( {1 + {{\left( {\log \,x} \right)}^2}} \right)}} - \frac{{2\,\log \,x}}{{{{\left( {1 + {{\left( {\log \,x} \right)}^2}} \right)}^2}}}} \right]} dx \cr & = \int {\left[ {\frac{{{e^t}}}{{1 + {t^2}}} - \frac{{2t\,{e^t}}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt\,\, \cr & {\text{put }}\log \,x = t\,\, \Rightarrow dx = {e^{t\,}}dt \cr & = \int {{e^t}} \left[ {\frac{1}{{1 + {t^2}}} - \frac{{2t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt \cr & \left[ {{\text{which is of the form }}\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} } \right] \cr & = \frac{{{e^t}}}{{1 + {t^2}}} + C\,\,\,\, = \frac{x}{{1 + {{\left( {\log \,x} \right)}^2}}} + C \cr} $$

$$\eqalign{ & \int {\frac{{\left( {{x^2} - 1} \right)}}{{x\sqrt {{x^4} + 3{x^2} + 1} }}} dx \cr & = \int {\frac{{\left( {{x^2} - 1} \right)}}{{{x^2}\sqrt {{x^2} + 3 + \frac{1}{{{x^2}}}} }}} dx \cr & = \int {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{\sqrt {{{\left( {x + \frac{1}{x}} \right)}^2} + 1} }}dx} \cr & = \int {\frac{{dz}}{{\sqrt {{z^2} + 1} }}} \,\,\,\,\,\left[ {{\text{Putting }}x + \frac{1}{x} = z \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)dx = dz} \right] \cr & = \log \left| {z + \sqrt {{z^2} + 1} } \right| + C \cr & = \log \left| {x + \frac{1}{x} + \sqrt {{x^2} + \frac{1}{{{x^2}}} + 3} } \right| + C \cr} $$

$$\eqalign{ & I = \int {\frac{{{{\sec }^2}x}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{9}{2}}}}}dx} \cr & {\text{Let }}\sec \,x + \tan \,x = t \cr & \Rightarrow \sec \,x - \tan \,x = \frac{1}{t} \cr & \Rightarrow \sec \,x = \frac{1}{2}\left( {t + \frac{1}{t}} \right) \cr & {\text{Also }}\sec x\left( {\sec \,x + \tan \,x} \right)dx = dt \cr & \Rightarrow \sec \,x\,dx = \frac{{dt}}{t} \cr & \therefore I = \frac{1}{2}\int {\frac{{\left( {t + \frac{1}{t}} \right)dt}}{{{t^{\frac{9}{2}}}.t}}} \cr & = \frac{1}{2}\int {\left( {{t^{ - \,\frac{9}{2}}} + {t^{ - \,\frac{{13}}{2}}}} \right)} dt \cr & = \frac{1}{2}\left[ {\frac{{{t^{ - \,\frac{9}{2}\, + \,1}}\,}}{{ - \frac{9}{2} + 1}} + \frac{{{t^{ - \,\frac{{13}}{2}\, + \,1}}}}{{ - \frac{{13}}{2} + 1}}} \right] + K \cr & = \frac{{ - 1}}{7}{t^{ - \,\frac{7}{2}}} - \frac{1}{{11}}{t^{ - \,\frac{{11}}{2}}} + K \cr & = - \frac{1}{{7{t^{\,\frac{7}{2}}}}} - \frac{1}{{11{t^{\,\frac{{11}}{2}}}}} + K \cr & = - \frac{1}{{{t^{\frac{{11}}{2}}}}}\left( {\frac{1}{{11}} + \frac{{{t^2}}}{7}} \right) + K \cr & = \frac{{ - 1}}{{{{\left( {\sec \,x + \tan \,x} \right)}^{\frac{{11}}{2}}}}}\left\{ {\frac{1}{{11}} + \frac{1}{7}{{\left( {\sec \,x + \tan \,x} \right)}^2}} \right\} + K \cr} $$

$$\eqalign{ & {\text{Put }}{x^{\frac{7}{2}}} = t \cr & \therefore \frac{7}{2}{x^{\frac{5}{2}}}dx = dt \cr & \therefore I = \int {\frac{2}{7}\frac{{dt}}{{\sqrt {1 + {t^2}} }}} \cr & = \frac{2}{7}\log \left( {t + \sqrt {1 + {t^2}} } \right) + k \cr & = \frac{2}{7}\log \left( {{x^{\frac{7}{2}}} + \sqrt {1 + {x^7}} } \right) + k \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {{e^{\ln \,x}}\sin \,x\,dx} \cr & = \int {x\,\sin \,x\,dx} \,\,\,\,\,\,\,\,\left( {\because \,{e^{\log \,a}} = a} \right) \cr & = - x\,\cos \,x + \int {1.\cos \,x\,dx} \cr & = \left( {\sin \,x - x\,\cos \,x} \right) + c \cr} $$

$$\eqalign{ & \int {{e^{3\,\log \,x}}{{\left( {{x^4} + 1} \right)}^{ - 1}}dx} \cr & = \int {{e^{\log \,{x^3}}}} \frac{1}{{{x^4} + 1}}dx \cr & = \int {\frac{{{x^3}}}{{{x^4} + 1}}dx} \cr & = \frac{1}{4}\log \left( {{x^4} + 1} \right) + C \cr & \left[ {{\text{since, }}{e^{{{\log }_e}{x^3}}} = {x^3}} \right] \cr} $$

$$I = \int {\frac{1}{{1 + 3\,{{\sin }^2}x + 8\,{{\cos }^2}x}}} dx$$
Dividing the numerator and denominator by $${\cos ^2}x,$$  we ge
$$\eqalign{ & I = \int {\frac{{{{\sec }^2}x}}{{{{\sec }^2}x + 3\,{{\tan }^2}x + 8}}} dx \cr & \Rightarrow I = \int {\frac{{{{\sec }^2}x}}{{1 + {{\tan }^2}x + 3\,{{\tan }^2}x + 8}}dx} \cr & \Rightarrow I = \int {\frac{{{{\sec }^2}x}}{{4\,{{\tan }^2}x + 9}}} dx \cr & {\text{Putting}}\,\,\tan \,x = t \Rightarrow {\sec ^2}x\,dx = dt,\,{\text{we get}} \cr & \Rightarrow I = \int {\frac{{dt}}{{4{t^2} + 9}}} \cr & \Rightarrow I = \frac{1}{4}\int {\frac{{dt}}{{{t^2} + {{\left( {\frac{3}{2}} \right)}^2}}}} \cr & \Rightarrow I = \frac{1}{4} \times \frac{1}{{\frac{3}{2}}}{\tan ^{ - 1}}\left( {\frac{t}{{\frac{3}{2}}}} \right) + C \cr & \Rightarrow I = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2t}}{3}} \right) + C \cr & \Rightarrow I = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2\,\tan \,x}}{3}} \right) + C \cr} $$

$$\eqalign{ & {\text{Let}}\,\,I = \sqrt 2 \int {\frac{{\sin \,xdx}}{{\sin \left( {x - \frac{\pi }{4}} \right)}}} \,\,{\text{put}}\,\,x - \frac{\pi }{4} = t \cr & \Rightarrow dx = dt\,\, \cr & \Rightarrow I = \sqrt 2 \int {\frac{{\sin \left( {t + \frac{\pi }{4}} \right)}}{{\sin \,t}}dt} \, \cr & \Rightarrow I = \frac{{\sqrt 2 }}{{\sqrt 2 }}\int {\left( {\frac{{\sin \,t + \cos \,t}}{{\sin \,t}}} \right)dt} \cr & \Rightarrow I = \int {\left( {1 + \cot \,t} \right)dt = t + \log \left| {\sin \,t} \right| + {c_1}} \cr & \Rightarrow I = x - \frac{\pi }{4} + \log \left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right| + {c_1} \cr & \Rightarrow I = x + \log \left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right| + c\,\,\left( {{\text{where}}\,\,c = {c_1} - \frac{\pi }{4}} \right) \cr} $$

$$\eqalign{ & {\text{Put }}x = \tan \,\theta \Rightarrow dx = {\sec ^2}\theta \,d\theta \cr & I = \int {{e^\theta }} \frac{{1 + \tan \,\theta + {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}.{\sec ^2}\theta \,d\theta \cr & \,\,\,\,\, = \int {{e^\theta }} \left( {\tan \,\theta + {{\sec }^2}\theta } \right)d\theta \cr & \,\,\,\,\, = {e^\theta }\tan \,\theta + c \cr & \,\,\,\,\, = x{e^{{{\tan }^{ - 1}}x}} + c \cr} $$